Problem: $h(r)=(r+1)(r+8)$ 1) What are the zeros of the function? Write the smaller $r$ first, and the larger $r$ second. $\text{smaller }r=$
Answer: $\begin{aligned} (r+1)(r+8)&=0 \\\\ r+1=0&\text{ or }r+8=0 \\\\ r={-1}&\text{ or }r={-8} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $r$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }r\text{-coordinate}&=\dfrac{({-1})+({-8})}{2} \\\\ &={-\dfrac{9}{2}} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $h\left({-\dfrac{9}{2}}\right)$ : $\begin{aligned} h\left({-\dfrac{9}{2}}\right)&=\left({-\dfrac{9}{2}}+1\right)\left({-\dfrac{9}{2}}+8\right) \\\\ &=\left(-\dfrac72\right)\left(\dfrac72\right) \\\\ &=-\dfrac{49}{4} \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }r&=-8 \\\\ \text{larger }r&=-1 \end{aligned}$ The vertex of the parabola is at $\left(-\dfrac{9}{2},-\dfrac{49}{4}\right)$